Could someone help me find the Covariance of these two distributions

[LHS]

Homework Statement

[PLAIN]http://img695.imageshack.us/img695/7551/unledsi.png

Homework Equations

The Attempt at a Solution

I get E=1/3 and E[V]=1, can't get E[UV] to be correct as I do not get the required answer, any help would be greatly appreciated! thanks!

 

[vela]

Your expected value of V sounds wrong. Both X and Y are distributed over the interval [0,1], so V=max{X,Y} would also be. Since the maximum value V can attain is 1 and it'll usually be less than 1, I'd think E[V]<1.

Show us what you've done so far so we can see where the problem lies.

 

[LHS]

Hmm.. yes, you are correct. I'm trying to work out E(U) now, then as max(X,Y)=1-min(X,Y) we can say E(V)=1-E(U)

But X and Y aren't independent so that's why I previously got it wrong.
So far have
P(U<u)=P(X<u,Y>u)+P(X<u,Y<u)+P(X>u,Y<u)

Not sure if that's correct, but previously I split these up and used the information about X and Y to find P(U<u). Kinda stuck seeing what to do here now..

 

[vela]

Hmm.. yes, you are correct. I'm trying to work out E(U) now, then as max(X,Y)=1-min(X,Y) we can say E(V)=1-E(U)

This isn't correct. For example, suppose X=1/4 and Y=1/2. Then min{X,Y}=1/4 and max{X,Y}=1/2, so max{X,Y} isn't equal to 1-min{X,Y}

But X and Y aren't independent so that's why I previously got it wrong.
So far have
P(U<u)=P(X<u,Y>u)+P(X<u,Y<u)+P(X>u,Y<u)

Not sure if that's correct, but previously I split these up and used the information about X and Y to find P(U<u). Kinda stuck seeing what to do here now..

 

[LHS]

Hmm.. you are right.. but surely it would still be true that E(V)=1-E(U)?

 

[vela]

Probably. It seems like it should from symmetry considerations.

What did you get for the P(U<u)? I found it equaled 2u.

 

[LHS]

So did I, I believe that must be correct. However I just said that it's because U must be uniformly distributed on [0,1/2] because Y=1-X and X,Y are uniformly distributed on [0,1], think there's a more rigorous way of proving this?

Also for E(UV), UV=min(X,Y)max(X,Y) so does this equal XY and hence E(UV)=E(X(1-X))
=E(X-X^2)
=E(X)-E(x^2)?

Edit: Yes, gives Cov(U,V)=-1/48 :)

 

[vela]

So did I, I believe that must be correct. However I just said that it's because U must be uniformly distributed on [0,1/2] because Y=1-X and X,Y are uniformly distributed on [0,1], think there's a more rigorous way of proving this?

If U<u, then you know that X<u or Y<u. In terms of just X, you have X<u or 1-X<u. So...

 

[LHS]

Ah of course! Thanks for your patience here, as you can probably tell probability is not my strong area.
If I can be cheeky enough to bother you anymore, any tips on going about finding the density of V/U?

 

[vela]

I've found probability to be one of the more difficult subjects as well. The basic concepts seem straightforward, but it's pretty easy to get confused trying to apply them correctly.

For V/U, try approaching it like you did with the covariance, so you get either X/Y or Y/X. It seems like it should work out.

 

[LHS]

Indeed, that's entirely the problem. Give me some calculus or dynamics any day!

Hmm.. so V/U equals Y/X or X/Y, so (1-X)/X or (1-Y)/Y, which both have the same distribution, so we can say:
P(V/U<a)=P((1-X)/X<a)=1-P(X<1/(a+1)) = a/(a+1)?

If not don't worry, I'll come back to it, Thank you ever so much! you've been extremely helpful!